Solve the integral $\int^{+\infty}_0\frac{x\sin x}{1+x^4}dx$ using complex analysis I am trying to solve the following integral using only complex analysis:
$$\int^{+\infty}_0\frac{x\sin x}{1+x^4}dx$$
So, as the function $f(x)$ inside the integral is even, the integral can be expressed this way:
$$\int^{+\infty}_0\frac{x\sin x}{1+x^4}dx = \frac{1}{2}\int^{+\infty}_{-\infty}\frac{x\sin x}{1+x^4}dx = I$$
Edit: If $x = Re (z)$, expressing $I$ in terms of $z$, I get:
$$ I = \int^{+\infty}_0\frac{ze^{iz}}{1+z^4}dz$$
$$f(z)= \frac{ze^{iz}}{1+z^4}$$
It has singularities on $z_0= \{\frac{1}{\sqrt{2}}(1+i),\frac{1}{\sqrt{2}}(1-i),\frac{1}{\sqrt{2}}(-1-i),\frac{1}{\sqrt{2}}(-1+i)\}$. If we take a contour $C$ with a upper-axis semi-circle B and the axis running from $−R→R$ and apply the residue theorem to the poles $z_0 = \{\frac{1}{\sqrt{2}}(1+i),\frac{1}{\sqrt{2}}(-1+i)\}$ I get
$$f(z)=\frac{ze^{iz}}{(z-(1+i)/\sqrt{2})(z-(1-i)/\sqrt{2})(z-(-1+i)/\sqrt{2})(z-(-1-i)/\sqrt{2})}$$
$$\int_C f(z)dz=2\pi i\sum_k Res(f,z_k)$$
Until this step, would this approach be correct?
 A: You're on the right track, but some points are in order.
Firstly, I don't think you meant $x=\Re z$; all you really need to switch to your chosen complex integrand is $\sin z=\Im e^{iz}$.
Secondly, we need to check the arc $z=Re^{i\theta}$ from $\theta=0$ to $\theta=\pi$ makes a contribution that $\to0$ as $R\to\infty$. It's enough to note that if $|R|\ge\sqrt[4]{2}$ then $|1+R^4e^{4i\theta}|\ge R^4-1\ge\tfrac12R^4$ by the triangle inequality, and$$\left|\int_0^\pi\frac{Re^{i\theta}e^{iRe^{i\theta}}iRe^{i\theta}d\theta}{1+R^4e^{4i\theta}}\right|\le \int_0^\pi\left|\frac{Re^{i\theta}e^{iRe^{i\theta}}iRe^{i\theta}}{\tfrac12R^4}\right|d\theta=2R^{-2}\int_0^\pi e^{-R\sin\theta}d\theta\le2\pi R^{-2}.$$
Thirdly, the slickest way to do the residue calculations notes that if $p^4+1=0$ then$$\lim_{z\to p}\frac{z(z-p)}{1+z^4}=\lim_{z\to p}\frac{2z-p}{4z^3}=\frac{1}{4p^2}$$by L'Hôpital's rule, so the residue of $\frac{ze^{iz}}{1+z^4}$ at $p$ is $\frac{e^{ip}}{4p^2}$. I leave you to do the rest yourself. (Just remember we only care about poles of positive imaginary part, which halves your work.) You should find $I=\tfrac{\pi}{2}\sin\tfrac{1}{\sqrt{2}}\exp\tfrac{-1}{\sqrt{2}}$.
